This is a first-order filter design problem.
Firstly, we need to find the transfer function from the Bode magnitude diagram, then
determine the cutoff frequency of the filter.
Given the template circuit in figure 2, we want to find the values of the components.
Let’s get started with part a, the transfer function.
For the low frequency range, it is approximately a straight line at around 5 dB.
It doesn’t start from 0 dB, so we know there is a constant K in the transfer function.
20 log10 k is 5 dB, so let’s try to get K.
Divide both sides by 20.
K should be 10 to the power of 0.25, or 1.778.
The first factor of the transfer function is 1.778.
For the high frequency range, it is a straight line with a slope of about 20 dB / dec.
The two lines meet at 1000 rad/s, so the cutoff frequency should be 1000 rad/s.
The curve goes downward by 20 dB, so it should be a simple pole located at 1000 rad/s.
The numerator and denominator are multiplied by 1000.
That’s the transfer function in frequency domain.
Let’s look at figure 1 and try to find the cutoff frequency.
The other name of cutoff frequency is 3 dB frequency.
That is, 3 dB below the pass band gain.
The pass band gain here is 5 dB, so we need to look at the 2 dB point.
When the frequency is 1000 rad/s, the magnitude is 2 dB, so the cutoff frequency should
be 1000 rad/s.
We need to design the filter based on the template given in figure 2.
Firstly, let’s try to find the transfer function of this circuit.
To make it general, let’s make the 10 kΩ resistor in s domain just R.
We need to convert the circuit into s domain.
The impedance of a capacitor is 1 / sC.
We use upper case V to indicate the variable in s domain.
Here, the transfer function is Vout / Vin.
We need to relate the two voltage variables to each other.
For the ideal op-amp, the non-inverting node is grounded, so that is 0 V.
The inverting node should have the same voltage as the non-inverting node, or 0 V.
We can write a Kirchhoff’s Current Law equation at the inverting node.
The current through Ri should be Vin – 0 / Ri.
The current through the 10 kΩ resistor should be Vout – 0 / R.
The voltage across the capacitor should be Vout / 1 / sC.
No current can flow into the op-amp through the input node.
We can rearrange the equation and find the transfer function.
Let’s write the simplified transfer function.
[equation]
In frequency domain, we get the transfer function by setting s equal to j ω.
[equation]
Let’s compare the template circuit transfer function with the transfer function we got
from the Bode diagram.
The gain in the circuit should be 1.778, so R / Ri should be 1.778.
The cutoff frequency is 1000 rad/s, so 1 / RC should be 1000 rad/s.
We know R is 10 kΩ in this circuit, so Ri should be R / 1.778.
We get a value of 5.6 kΩ for Ri.
Let’s look at the cutoff frequency.
Capacitance should be 1 / R 1000.
The resistance R is 10 kΩ, then times 1000 gives us 0.1 µF.
We have the values for the components.
Ri is 5.6 kΩ.
The capacitance is 0.1 µF.
This circuit should have a magnitude plot as given in figure 1.